This problem tests the understanding of array manipulation and basic arithmetic operations. It involves dealing with an integer array representing a large integer to which we have to add

Given an integer array called `nums`

that represents a large integer;** **our task is to increment this number and return the resulting integer array. Each element in the `nums`

array corresponds to a digit of the integer where digits are ordered from most significant to least significant in left-to-right order.

Note:The given large integer does not have any leading$0$ ’s.

**Constraints:**

$1\leq$ `nums.length`

$\leq 100$ $0\leq$ `nums[i]`

$\leq 9$

Let’s take a look at a few examples to get a better understanding of the problem statement:

Plus One LeetCode

1

Given the following inputs, what will be the output?

`nums`

= [1, 0]

A)

[11]

B)

[1, 1]

C)

[9]

D)

11

Question 1 of 20 attempted

To solve the given problem, we will traverse the array from the end to the beginning. We start by moving along the `nums`

array from the last digit. For each element, we check if it is a

Let’s look at the algorithm:

Traverse the

`nums`

array from the last element toward the first.If there are

$9$ ’s at the end of the array, change them to$0$ ’s.Upon encountering a digit that is not

$9$ , increment it by$1$ and return the array, as no further steps are needed.If all digits are

$9$ ’s, they will now be$0$ ’s. In this scenario, we prepend the digit$1$ to the array and return the result.

Let’s look at the following illustration to get a better understanding of the solution:

Let’s look at the code for the algorithm we just discussed.

def plus_one(nums):n = len(nums) - 1# Traversing the array from end to startfor i in range(n, -1, -1):# If the element of the nums array is 9, change it to 0if nums[i] == 9:nums[i] = 0# Otherwise, increment 1 and return the resultelse:nums[i] += 1return nums# Add 1 to the start of the array if all digits were 9return [1] + nums# Driver codedef main():nums = [[1, 0, 5], [9, 9], [1, 9, 3, 2, 1], [6, 8], [1, 7, 6, 9]]for i in range(len(nums)):print(i+1, '.', '\tGiven array: ', nums[i], sep='')result = plus_one(nums[i])print('\n\tThe result: ', result)print('-' * 100)if __name__ == '__main__':main()

Plus One

After discussing the solution to the given problem, we will now discuss its complexity analysis.

The time complexity of the solution is `nums`

array.

The space complexity is

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